Question

A \(2.20-\) g sample of an unknown acid (empirical formula \(=\) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3} )\) is dissolved in 1.0 \(\mathrm{L}\) of water. A titration required 25.0 mL of 0.500 M NaOH to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

Short answer

The molecular formula of the unknown acid is \(\mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}\).

Step-by-step solution

Find moles of NaOH used in the titration

To find the moles of NaOH used in the titration, we can use the formula: Moles of NaOH = volume of NaOH (L) × molarity of NaOH Given: Volume of NaOH = 25.0 mL = 0.025 L (conversion from mL to L) Molarity of NaOH = 0.500 M Moles of NaOH = (0.025 L) × (0.500 mol/L) = 0.0125 mol

Find moles of the unknown acid

Since we know that 1 mole of the acid reacts with 1 mole of NaOH, the moles of the unknown acid are equal to the moles of NaOH: Moles of unknown acid = Moles of NaOH = 0.0125 mol

Find the molar mass of the acid

Now that we know the moles of the unknown acid, we can use the mass of the sample to find the molar mass. Molar mass of the unknown acid = mass of the unknown acid (g) / moles of the unknown acid Given: Mass of the unknown acid = 2.20 g Moles of the unknown acid = 0.0125 mol Molar mass of the unknown acid = 2.20 g / 0.0125 mol = 176 g/mol

Find the ratio of the empirical formula molar mass to the molar mass

We need to calculate the ratio of the empirical formula's molar mass (\(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3}\)) to the molar mass found in step 3: Empirical formula molar mass: 3 × (12 g/mol for C) + 4 × (1 g/mol for H) + 3 × (16 g/mol for O) = 88 g/mol Ratio = (Molar mass of the unknown acid) / (Empirical formula molar mass) = 176 g/mol / 88 g/mol = 2

Determine the molecular formula

Now we multiply the empirical formula by the ratio found in step 4 to determine the molecular formula: Molecular formula = \(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3} × 2 = \mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}\) The molecular formula of the unknown acid is \(\mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}\).

Key concepts

Moles

Moles are a fundamental concept in chemistry, providing a bridge between the atomic scale and the world we interact with daily. Essentially, a mole is a unit that allows chemists to quantify the amount of substance. If you imagine trying to count atoms, the numbers would be mind-bogglingly large, so the mole acts as a convenient grouping. One mole contains Avogadro's number of entities, approximately \(6.022 \times 10^{23}\). This number represents the count of atoms in 12 grams of carbon-12.

In our titration exercise, we determine the number of moles of NaOH used in the reaction. We used the volume and molarity of NaOH to find this number. The formula used is:

• \(\text{Moles of NaOH} = \text{Volume of NaOH (L)} \times \text{Molarity of NaOH}\)

By finding that 0.0125 moles of NaOH were used, we indirectly determine the moles of the unknown acid, as one mole of NaOH reacts with one mole of the acid.

Molarity

Molarity is a measure of concentration used for solutions in chemistry. It is defined as the number of moles of a solute present in one liter of solution. The formula is straightforward:

• \(\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\)

This is handy because it allows chemists to prepare solutions with precise concentrations, crucial for reactions like titrations.

In our exercise, the molarity of NaOH was used to determine how much of the acid it could neutralize. Given a 0.500 M NaOH solution, we calculated how many moles of NaOH were present in 25 mL using molarity, helping us connect the titrant's concentration to the unknown acid sample.

Empirical Formula

The empirical formula of a compound provides the smallest whole number ratio of the atoms of each element present. It gives us a simplified picture of the compound's composition. This formula is often derived from experimental data regarding the relative masses of each element in a compound.

In the problem, the empirical formula \(\text{C}_3\text{H}_4\text{O}_3\) suggests how many carbon, hydrogen, and oxygen atoms are present per molecule, in the simplest whole-number terms. Still, it does not tell us the actual number of atoms in a molecule unless the compound has exactly one empirical formula per molecule.

To progress from this formula to the molecular formula, we need an additional step wherein we calculate the molecular weight and compare this to the empirical formula weight.

Molecular Formula

The molecular formula reveals the exact number of each type of atom in a single molecule of a compound, offering more detailed insight than the empirical formula. While the empirical formula lays down a ratio, the molecular formula shows the actual numbers.

To find the molecular formula, one must first determine the molar mass of the compound. After calculating the moles and molar mass of the unknown acid as 176 g/mol, we use the ratio of this molar mass to the empirical formula molar mass (88 g/mol for \((\text{C}_3\text{H}_4\text{O}_3)\)).

The ratio found was 2, so the empirical formula is multiplied by 2 to give the molecular formula \((\text{C}_6\text{H}_8\text{O}_6)\). This tells us that each molecule contains six carbon atoms, eight hydrogen atoms, and six oxygen atoms.